Question

Let ø(x, y) = arctan (y/3x).

(a) Find the vector field F := Vø, and simplify its components as much as possible.

(b) What is the domain of definition of F?

(c) Find all the points (a, b) (belonging to the domain of definition

F) at which the vector F(a,b) is parallel to the line y = x.

of

(Note: For a real number u, arctan u is also denoted by tan -1 u. Use

the notation that you prefer.)

Answer 1

SOLUTION

GIVEN

$\displaystyle \sf{ \phi(x,y) = arc \tan \bigg( \frac{y}{3x} \bigg)}$

TO DETERMINE

(a) Find the vector field F := Vø, and simplify its components as much as possible.

(b) What is the domain of definition of F?

(c) Find all the points (a, b) (belonging to the domain of definition ) at which the vector F(a,b) is parallel to the line y = x.

EVALUATION

Here it is given that

$\displaystyle \sf{ \phi(x,y) = arc \tan \bigg( \frac{y}{3x} \bigg) ={\tan}^{ - 1} \bigg( \frac{y}{3x} \bigg) }$

a) Now

$\displaystyle \sf{ \frac{ \partial \phi}{ \partial x} }$

$\displaystyle \sf{ = \frac{1}{1 +{ \big( \frac{y}{3x} \big)}^{2} } \times \frac{ -y}{3 {x}^{2} } }$

$\displaystyle \sf{ = \frac{9 {x}^{2} }{9 {x}^{2} + {y}^{2} } \times \frac{ -y}{3 {x}^{2} } }$

$\displaystyle \sf{ = \frac{ - 3y }{9 {x}^{2} + {y}^{2} } }$

Again

$\displaystyle \sf{ \frac{ \partial \phi}{ \partial y} }$

$\displaystyle \sf{ = \frac{1}{1 +{ \big( \frac{y}{3x} \big)}^{2} } \times \frac{1}{3 x } }$

$\displaystyle \sf{ = \frac{9 {x}^{2} }{9{x}^{2} + {y}^{2} } \times \frac{1}{3 x } }$

$\displaystyle \sf{ = \frac{3x }{9{x}^{2} + {y}^{2} } }$

Now

$\displaystyle \vec{F} = \nabla \phi = \frac{ \partial \phi}{ \partial x} \hat{i} + \frac{ \partial \phi}{ \partial y} \hat{j}$

Thus the required component form is

$\displaystyle \sf{ = \bigg( \: \frac{ \partial \phi}{ \partial x} \: , \: \frac{ \partial \phi}{ \partial y} \: \bigg)}$

$\displaystyle \sf{ = \bigg( \: \frac{ - 3y }{9 {x}^{2} + {y}^{2} } \: , \: \frac{ 3x }{9 {x}^{2} + {y}^{2} } \: \bigg)}$

b) We observe that F is not well defined when denominator of the components vanishes

Which occurs only when

$\sf{9 {x}^{2} + {y}^{2} = 0}$

$\implies \sf{x = 0 \: \: and \: y = 0 }$

Hence the required Domain of definition is

${ \mathbb {R}^{2} - (0,0)}$

c) For determining the the points (a, b) (belonging to the domain of definition ) at which the vector F(a,b) is parallel to the line y = x we have

Abscissa = ordinate

$\displaystyle \sf{ \implies \: \frac{ - 3y }{9 {x}^{2} + {y}^{2} } \: = \: \frac{ 3x }{9 {x}^{2} + {y}^{2} } }$

$\displaystyle \sf{ \implies \:y = - x }$

Thus the points are of the form

(x, - x) where x ≠ 0

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